Fdtxjr

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5

$begingroup$

Here are two locations in memory:

 0110 | 1111 1110 1101 0011
 0111 | 0000 0110 1101 1001

Interpret locations 6 (0110) and 7 (0111) as an IEEE floating point number.
Location 6 contains bits [15:0] and location 7 contains bits [16:31].

Floating Point -
 Locations 6 and 7: 0000 0110 1101 1001 1111 1110 1101 0011
 The number represented is 1.101 1001 1111 1110 1101 0011 × 2^(-114) 

Is my answer correct?

I am unsure exactly what the [16:31] and [15:0] relate to/mean?

floating-point

share|cite|improve this question

edited Mar 14 at 7:13

Winfield Chen

484

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not really a math question. You may have better luck on a computing site.
  $endgroup$
  – Gerry Myerson
  Sep 14 '14 at 7:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson Can't we migrate this for him? It's been done for at least one my questions and maybe one of yours.
  $endgroup$
  – Robert Soupe
  Sep 14 '14 at 17:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert, I suppose we could vote to close and migrate, or flag for moderator attention and suggest migration, but I don't know enough about computing sites to suggest one as the target.
  $endgroup$
  – Gerry Myerson
  Sep 15 '14 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gerry, I was looking at programmers.stackexchange, but a question there about converting from single precision float to half precision was closed for being "off-topic" (which of course does not always mean it really is off-topic. The scope is said to include questions about "algorithm and data structure concepts."
  $endgroup$
  – Robert Soupe
  Sep 15 '14 at 2:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert, if you find an appropriate site, go ahead and take whatever steps you can to migrate. I'm happy to let OP do the work of finding a better site and the work of contacting the moderators to migrate, or let OP delete here and post elsewhere.
  $endgroup$
  – Gerry Myerson
  Sep 15 '14 at 3:53
  

  
  
  
  

add a comment | 

5

$begingroup$

Here are two locations in memory:

 0110 | 1111 1110 1101 0011
 0111 | 0000 0110 1101 1001

Interpret locations 6 (0110) and 7 (0111) as an IEEE floating point number.
Location 6 contains bits [15:0] and location 7 contains bits [16:31].

Floating Point -
 Locations 6 and 7: 0000 0110 1101 1001 1111 1110 1101 0011
 The number represented is 1.101 1001 1111 1110 1101 0011 × 2^(-114) 

Is my answer correct?

I am unsure exactly what the [16:31] and [15:0] relate to/mean?

floating-point

share|cite|improve this question

edited Mar 14 at 7:13

Winfield Chen

484

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not really a math question. You may have better luck on a computing site.
  $endgroup$
  – Gerry Myerson
  Sep 14 '14 at 7:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson Can't we migrate this for him? It's been done for at least one my questions and maybe one of yours.
  $endgroup$
  – Robert Soupe
  Sep 14 '14 at 17:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert, I suppose we could vote to close and migrate, or flag for moderator attention and suggest migration, but I don't know enough about computing sites to suggest one as the target.
  $endgroup$
  – Gerry Myerson
  Sep 15 '14 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gerry, I was looking at programmers.stackexchange, but a question there about converting from single precision float to half precision was closed for being "off-topic" (which of course does not always mean it really is off-topic. The scope is said to include questions about "algorithm and data structure concepts."
  $endgroup$
  – Robert Soupe
  Sep 15 '14 at 2:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert, if you find an appropriate site, go ahead and take whatever steps you can to migrate. I'm happy to let OP do the work of finding a better site and the work of contacting the moderators to migrate, or let OP delete here and post elsewhere.
  $endgroup$
  – Gerry Myerson
  Sep 15 '14 at 3:53
  

  
  
  
  

add a comment | 

$begingroup$

Here are two locations in memory:

 0110 | 1111 1110 1101 0011
 0111 | 0000 0110 1101 1001

Interpret locations 6 (0110) and 7 (0111) as an IEEE floating point number.
Location 6 contains bits [15:0] and location 7 contains bits [16:31].

Floating Point -
 Locations 6 and 7: 0000 0110 1101 1001 1111 1110 1101 0011
 The number represented is 1.101 1001 1111 1110 1101 0011 × 2^(-114) 

Is my answer correct?

I am unsure exactly what the [16:31] and [15:0] relate to/mean?

floating-point

share|cite|improve this question

edited Mar 14 at 7:13

Winfield Chen

484

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

$endgroup$

Floating Point -
 Locations 6 and 7: 0000 0110 1101 1001 1111 1110 1101 0011
 The number represented is 1.101 1001 1111 1110 1101 0011 × 2^(-114) 

share|cite|improve this question

edited Mar 14 at 7:13

Winfield Chen

484

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

share|cite|improve this question

edited Mar 14 at 7:13

Winfield Chen

484

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

edited Mar 14 at 7:13

Winfield Chen

484

edited Mar 14 at 7:13

Winfield Chen

484

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

asked Sep 14 '14 at 6:03

Patrick StevensPatrick Stevens

61113

1 Answer
1

active

oldest

votes

4

+25

$begingroup$

The [16:31] and [15:0] refer to locations in the binary representation of a $32$-bit integer. You have interpreted this correctly.

When in doubt about technical problems, always consult Wikipedia an expert.

In your case the number is
0 || 00001101 || 101 1001 1111 1110 1101 0011

The sign is positive.

The biased exponent is 1101 $ = 13$, so the actual exponent is $13 - 127 = -114$, assuming single precision.

So the answer you have is correct:
$$2^-114 times (1.101 1001 1111 1110 1101 0011)_2$$

share|cite|improve this answer

edited Sep 17 '14 at 18:17

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If this question didn't advocate trusting Wikipedia at any level on any topic, and if I was the issuer of the bounty, I think I would award it to this answer.
  $endgroup$
  – Robert Soupe
  Sep 17 '14 at 0:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd let the typo slide, though ("techincal").
  $endgroup$
  – Robert Soupe
  Sep 17 '14 at 1:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hows that? Better?
  $endgroup$
  – A.E
  Sep 17 '14 at 2:38
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Endianness do matter. You can check here
  $endgroup$
  – user137035
  Sep 17 '14 at 10:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Agreed, endianness matters. I am assuming that the bits in memory correspond to a 32 bit word (in the order stated) and that the exponent and mantissa have the MSB on the left.
  $endgroup$
  – A.E
  Sep 17 '14 at 18:24
  

  
  
  
  

add a comment | 

Your Answer

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4

+25

$begingroup$

The [16:31] and [15:0] refer to locations in the binary representation of a $32$-bit integer. You have interpreted this correctly.

When in doubt about technical problems, always consult Wikipedia an expert.

In your case the number is
0 || 00001101 || 101 1001 1111 1110 1101 0011

The sign is positive.

The biased exponent is 1101 $ = 13$, so the actual exponent is $13 - 127 = -114$, assuming single precision.

So the answer you have is correct:
$$2^-114 times (1.101 1001 1111 1110 1101 0011)_2$$

share|cite|improve this answer

edited Sep 17 '14 at 18:17

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If this question didn't advocate trusting Wikipedia at any level on any topic, and if I was the issuer of the bounty, I think I would award it to this answer.
  $endgroup$
  – Robert Soupe
  Sep 17 '14 at 0:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd let the typo slide, though ("techincal").
  $endgroup$
  – Robert Soupe
  Sep 17 '14 at 1:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hows that? Better?
  $endgroup$
  – A.E
  Sep 17 '14 at 2:38
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Endianness do matter. You can check here
  $endgroup$
  – user137035
  Sep 17 '14 at 10:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Agreed, endianness matters. I am assuming that the bits in memory correspond to a 32 bit word (in the order stated) and that the exponent and mantissa have the MSB on the left.
  $endgroup$
  – A.E
  Sep 17 '14 at 18:24
  

  
  
  
  

add a comment | 

4

+25

$begingroup$

The [16:31] and [15:0] refer to locations in the binary representation of a $32$-bit integer. You have interpreted this correctly.

When in doubt about technical problems, always consult Wikipedia an expert.

In your case the number is
0 || 00001101 || 101 1001 1111 1110 1101 0011

The sign is positive.

The biased exponent is 1101 $ = 13$, so the actual exponent is $13 - 127 = -114$, assuming single precision.

So the answer you have is correct:
$$2^-114 times (1.101 1001 1111 1110 1101 0011)_2$$

share|cite|improve this answer

edited Sep 17 '14 at 18:17

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If this question didn't advocate trusting Wikipedia at any level on any topic, and if I was the issuer of the bounty, I think I would award it to this answer.
  $endgroup$
  – Robert Soupe
  Sep 17 '14 at 0:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd let the typo slide, though ("techincal").
  $endgroup$
  – Robert Soupe
  Sep 17 '14 at 1:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hows that? Better?
  $endgroup$
  – A.E
  Sep 17 '14 at 2:38
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Endianness do matter. You can check here
  $endgroup$
  – user137035
  Sep 17 '14 at 10:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Agreed, endianness matters. I am assuming that the bits in memory correspond to a 32 bit word (in the order stated) and that the exponent and mantissa have the MSB on the left.
  $endgroup$
  – A.E
  Sep 17 '14 at 18:24
  

  
  
  
  

add a comment | 

$begingroup$

The [16:31] and [15:0] refer to locations in the binary representation of a $32$-bit integer. You have interpreted this correctly.

When in doubt about technical problems, always consult Wikipedia an expert.

In your case the number is
0 || 00001101 || 101 1001 1111 1110 1101 0011

The sign is positive.

The biased exponent is 1101 $ = 13$, so the actual exponent is $13 - 127 = -114$, assuming single precision.

So the answer you have is correct:
$$2^-114 times (1.101 1001 1111 1110 1101 0011)_2$$

share|cite|improve this answer

edited Sep 17 '14 at 18:17

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722

$endgroup$

share|cite|improve this answer

edited Sep 17 '14 at 18:17

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722

answered Sep 16 '14 at 21:29

A.EA.E

1,7101722